Prove the geometric sequence $(r^n)$ is Cauchy if $|r|<1$

Question

Suppose, towards a contradiction, that $(r^n)$ is not Cauchy. Then $\exists \epsilon >0$ such that for every $n\in \mathbb{N}$, $\exists m > n$ such that $|r^m - r^n| \geq \epsilon$. Then $|r^n| > |r^n(r^{m-n} - 1)| = |r^m - r^n| \geq \epsilon$.

This is as far as I got assuming only that the sequence is bounded by -1 and 1, which I'm not even sure if I'm allowed to assume.

The question is at the end of a chapter that goes over Cauchy sequences, and how they allow us to define products of reals by showing that for $x,y\in \mathbb{R}$, $\lim_{n\rightarrow \infty} (x_ny_n)$ converges, where $(x_n), (y_n)\subset \mathbb{Q}$ converge to $x,y$ respectively.

Any help would be appreciated.

Edit: Since $|r|< 1$, $|r|= (1 - k)$ for some $0<k<1$, so $r^{n+1} = r^n(1-k) < r^n$, and the sequence is strictly decreasing. Since $(|r^n|)$ is bounded below by 0, the sequence must converge, so the sequence is Cauchy.

Answer 1

You know that every convergent sequence is Cauchy.

You also know that the geometric sequence with $|r|<1$ is convergent.

Therefore the geometric sequence with $|r|<1$ is Cauchy.

Answer 2

Note that if $n >m$ then $|r|^n = |r|^{n-m} |r|^m < |r|^m$.

Let $\epsilon>0$ and suppose $N$ is such that $(1+|r|)|r|^N < \epsilon$. Then if $m,n \ge N$ we have (assuming that $n\ge m$ without loss of generality) that $|r^n-r^m| = |r|^m||r^{n-m}-1| \le |r|^m (1+|r|) \le (1+|r|)|r|^N < \epsilon$.

Answer 3

We have, with $m \ge n$,

$\vert r^n - r^m \vert = \vert r^n(1 - r^{m - n}) \vert = \vert r^n \vert \vert 1 - r^{m - n} \vert = \vert r \vert^n \vert 1 - r^{m - n} \vert; \tag 1$

also, again with $m \ge n$,

$\vert 1 - r^{m - n} \vert \le \vert 1 \vert + \vert r^{m - n} \vert = 1 + \vert r \vert^{m - n} \le 2; \tag 2$

we combine (1) and (2) and find

$\vert r^n - r^m \vert \le 2\vert r \vert^n; \tag 3$

now with $n$ sufficiently large we have

$2 \vert r \vert^n < \epsilon \tag 4$

for any $0 < \epsilon \in \Bbb R$; in fact, (4) is obtained when

$\ln 2 + n \ln \vert r \vert < \ln \epsilon, \tag 5$

or

$n \ln \vert r \vert < \ln \epsilon - \ln 2 = \ln \left ( \dfrac{\epsilon}{2} \right ), \tag 6$

or, since $\vert r \vert < 1$ implies $\ln \vert r \vert < 0$,

$n > (\ln \epsilon - \ln 2) / \ln \vert r \vert = \ln \left ( \dfrac{\epsilon}{2} \right ) / \ln \vert r \vert; \tag 6$

thus, given $\epsilon$, taking $N \in \Bbb N$ such that

$N > \ln \left ( \dfrac{\epsilon}{2} \right ) / \ln \vert r \vert; \tag 7$

we find that

$m \ge n > N \Longrightarrow \vert r^n - r^m \vert < \epsilon, \tag 8$

that is, that $r^n$ is Cauchy.

Actually, our OP hiroshin's approach isn't all that far off, since by taking $n$ large enough we will obtain $\vert r \vert^n < \epsilon$, a decisive contradiction.